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Math Help - Differential equations substitution

  1. #1
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    Differential equations substitution

    Solve:

    2y'y''=1+(y')^2 by letting y'=v

    Does y''=v'

    So does 2vv'=1+v^2
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  2. #2
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    Right. So what can you do now?
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  3. #3
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    What is v' with respect to?
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  4. #4
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    The same variable with respect to which you are taking the derivative of y. Your problem mentions no independent variable - I would probably just go with x.
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  5. #5
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    Quote Originally Posted by Ackbeet View Post
    The same variable with respect to which you are taking the derivative of y. Your problem mentions no independent variable - I would probably just go with x.
    So, \dfrac{2v}{1+v^2} \ dv=dx

    \displaystyle\int{\dfrac{2v}{1+v^2} \ dv}=\int{dx}

    \ln|1+v^2|=x+C

    1+v^2=Ce^x

    v^2=Ce^x-1
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  6. #6
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    Not sure I would use the division symbol there. On the LHS, you have dv/dx, originally. When you multiply both sides of

    \displaystyle{\frac{2v}{1+v^{2}}\,\frac{dv}{dx}=1} by dx, what are you going to get?

    [EDIT]: acevipa edited post # 5. It's correct now.
    Last edited by Ackbeet; September 3rd 2010 at 03:56 AM.
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  7. #7
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    Didn't I do it correct?
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  8. #8
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    Oh, I see. You edited your post. That looks good to me.
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  9. #9
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    Quote Originally Posted by Ackbeet View Post
    Oh, I see. You edited your post. That looks good to me.
    Therefore, (y')^2=Ce^x-1

    What do you do from here? Don't we have to find y?
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